Trying to be explicit.

Substitute $x=u^{-1}v^{-\frac{b_1}{a_2}}$, $y=v^{-1}u^{-\frac{b_2}{a_1}}$ (here $b_1=b_2=b$ but they could as well be different).

We obtain $$ \frac{du}{dt}=a_1v^{-\frac{b_1}{a_2}},\qquad\frac{dv}{dt}=a_2u^{-\frac{b_2}{a_1}}, $$ which is equivalent to $$ a_2u^{-\frac{b_2}{a_1}}du=a_1v^{-\frac{b_1}{a_2}}dv. $$ If $b_2\ne a_1$ and $b_1\ne a_2$ the general solution is given by $$ (a_2-b_1)u^{1-\frac{b_2}{a_1}}-(a_1-b_2)v^{1-\frac{b_1}{a_2}}=\mathrm{const.} $$ In case $b_2=a_1$ or $b_1=a_2$ or both we get logarithms instead of powers at appropriate places.

Trying to be explicit.

Substitute $x=u^{-a_1}v^{-b_1}$, $y=v^{-a_2}u^{-b_2}$ (here $b_1=b_2=b$ but they could as well be different).

We obtain $$ \frac{du}{dt}=u^{1-a_1}v^{-b_1},\qquad\frac{dv}{dt}=v^{1-a_2}u^{-b_2}, $$ which is equivalent to $$ u^{a_1-b_2-1}du=v^{a_2-b_1-1}dv. $$ If $b_2\ne a_1$ and $b_1\ne a_2$ the general solution is given by $$ \frac{u^{a_1-b_2}}{a_1-b_2}-\frac{v^{a_2-b_1}}{a_2-b_1}=\mathrm{const.} $$ In case $b_2=a_1$ or $b_1=a_2$ or both we get logarithms instead of powers at appropriate places.